2/23/16 - Review of Charge, Voltage, and Current plus Breadboards

Here we define current I to be the rate of change of charge q with respect to time t. Given a charge function in terms of t q(t) we can differentiate it to find a current function in terms of t I(t).

Since we know that current is the rate of change of charge q over time t, we can integrate to find the total charge over a given current from an amount of time. We're given an exponentially decreasing current i(t) = exp(-2t) and we integrate from 0 to 2 seconds we found that the total charge from the current to be 0.491C.

Here we are given a graph of an current function i(t), and are asked to sketch the graph of the power and total energy over time. We start with defining power P to be voltage V times current I and since our voltage is constant, we found that the power function P(t) graph is simply a scaled version of the current graph. Next, we define total energy E to be the area under the power graph with respect to time P(t). Since we have a constant piece wise power function, we find a linear graph for total energy over time E(t).

Our activity for the day was to explore the connections of a breadboard, a plastic board with slots to connect electrical components to a temporary circuit board. We explore the different resistances of the breadboard to see how it is connected beneath the pins.

First case is to connect two in the same row, we see that there is around 2.7 Ohms of resistance.

Next we test two different holes on separate columns, we observe that there is an infinite amount of resistance according to the millimeter, and concluded that there is no connection between these holes.

Here we connected two arbitrary holes, infinite resistance again so no connection.

Two holes on the other column, same row, return 2.9 Ohms. We conclude there is a connection between these two.

Our conclusions in a table.

Professor Mason's example of a breadboard with the voltage rails on the side.

Here we are given the graphs and functions of voltage and current with respect to time V(t) and i(t). We shape out the power graph P(t) as a quadratic where both voltage and current are linear and power as a linear function when either current or voltage is constant. Then we find the total energy function E(t) as an integral of power. Since the power is quadratic and linear, the energy graph turns into a cubic and quadratic, we then solve for the actual values and functions.

Here we are given a simple circuit loop and are asked to find the voltage across a circuit element given every other circuit element's voltages and currents. Since the power in the loop has to be dissipated from start to finish, and we're given the current and voltage out of the source, we can find the power that the element has to consume to make the entire circuit work.